use the Stefan–Boltzmann law L = 4πσr 2 T

Stellar Radii – Using the Stefan–Boltzmann Law 🌞

1. The Stefan–Boltzmann Law

The law states that the power radiated by a perfect black‑body is proportional to the fourth power of its surface temperature: $$L = 4\pi\sigma r^2 T^4$$ where:

• $L$ = luminosity (Watts)
• $\sigma$ = Stefan–Boltzmann constant ≈ $5.67\times10^{-8}\,\text{W m}^{-2}\text{K}^{-4}$
• $r$ = radius (metres)
• $T$ = effective surface temperature (Kelvin)

2. What Does It Mean?

Think of a star like a giant glowing lamp. The brighter the lamp (higher $L$) and the hotter its surface (higher $T$), the larger it must be to emit that power. If you keep the temperature the same but increase the radius, the star’s surface area grows, so it can shine brighter.

3. Solving for the Stellar Radius

Rearranging the equation to isolate $r$ gives: $$r = \sqrt{\frac{L}{4\pi\sigma T^4}}$$ Key points:

• Always keep units consistent – use SI units (Watts, metres, Kelvin).
• Because $T$ is raised to the fourth power, small errors in temperature can lead to large errors in radius.

4. Step‑by‑Step Example

1. Suppose a star has a luminosity $L = 3.828\times10^{26}\,\text{W}$ (same as the Sun) and a surface temperature $T = 5778\,\text{K}$.
2. Plug into the radius formula: $$r = \sqrt{\frac{3.828\times10^{26}}{4\pi(5.67\times10^{-8})(5778)^4}}$$
3. Compute the denominator first:
   • $4\pi\sigma \approx 4\pi \times 5.67\times10^{-8} \approx 7.13\times10^{-7}$
   • $T^4 = 5778^4 \approx 1.11\times10^{14}$
   • Denominator ≈ $7.13\times10^{-7} \times 1.11\times10^{14} \approx 7.92\times10^7$
4. Now divide $L$ by the denominator: $$\frac{3.828\times10^{26}}{7.92\times10^7} \approx 4.83\times10^{18}$$
5. Take the square root: $$r \approx \sqrt{4.83\times10^{18}} \approx 6.95\times10^8\,\text{m}$$
6. Compare to the Sun’s known radius $6.96\times10^8\,\text{m}$ – the calculation checks out! 🌟

5. Quick Reference Table

6. Exam Tips & Tricks